AP Biology Practice Exam 2 - Build Your Test-Taking Confidence - 5 Steps to a 5: AP Biology 2017 (2016)

5 Steps to a 5: AP Biology 2017 (2016)


Build Your Test-Taking Confidence

AP Biology Practice Exam 1

Answer Sheet for AP Biology Practice Exam 2





AP Biology Practice Exam 2: Section I


Time–1 hour and 30 minutes (for Parts A and B)

For the multiple-choice questions to follow, select the best answer and fill in the appropriate letter on the answer sheet.

1 . A baby duck runs for cover when a large object is tossed over its head. After this object is repeatedly passed overhead, the duck learns there is no danger and stops running for cover when the same object appears again. This is an example of

A. imprinting.

B. fixed-action pattern.

C. agonistic behavior.

D. habituation.

2 . In a population of giraffes, an environmental change occurs that favors individuals that are tallest. As a result, more of the taller individuals are able to obtain nutrients and survive to pass along their genetic information. This is an example of

A. directional selection.

B. stabilizing selection.

C. sexual selection.

D. disruptive selection.

3 . The relatives of a group of pelicans from the same species that separated from each other because of an unsuccessful migration are reunited 150 years later and find that they are unable to produce offspring. This is an example of

A. allopatric speciation.

B. sympatric speciation.

C. genetic drift.

D. gene flow.

E. natural selection.

4 . A cell is placed into a hypertonic environment and its cytoplasm shrivels up. This demonstrates the principle of

A. diffusion.

B. active transport.

C. facilitated diffusion.

D. plasmolysis.

5 . Which of the following is a biotic factor that could affect the growth rate of a population?

A. Volcanic eruption

B. Glacier melting

C. Destruction of the ozone layer

D. Sudden reduction in the animal food resource

6 . Which of the following is not a way to form recombinant DNA?

A. Translation

B. Conjugation

C. Specialized transduction

D. Transformation

7 . Chemiosmosis occurs in

I. Mitochondria

II. Nuclei

III. Chloroplasts

A. I only

B. II only

C. III only

D. I and III

8 . Which of the following theories is based on the notion that mitochondria and chloroplasts evolved from prokaryotic cells?

A. Fluid mosaic model

B. Endosymbiotic model

C. Taxonomic model

D. Respiration feedback model

9 . Which of the following is not known to be involved in the control of cell division?

A. Cyclins

B. Protein kinases

C. Checkpoints

D. Fibroblast cells

10 . Which of the following statements about post-transcriptional modification are incorrect?

A. A poly-A tail is added to the 3′ end of the mRNA.

B. A guanine cap is added to the 5′ end of the mRNA.

C. Introns are removed from the mRNA.

D. Posttranscriptional modification occurs in the cytoplasm.

11 . In a certain pond, there are long-finned fish and short-finned fish. A horrific summer thunderstorm leads to the death of a disproportionate number of long-finned fish to the point where the relative frequency of the two forms has drastically shifted. This is an example of

A. gene flow.

B. natural selection.

C. genetic drift.

D. stabilizing selection.

12 . Which of the following cells is most closely associated with phagocytosis?

A. Neutrophils

B. Plasma cells

C. B cells

D. Memory cells

13 . Which of the following statements about photosynthesis is incorrect ?

A. H2 O is an input to the light-dependent reactions.

B. CO2 is an input to the Calvin cycle.

C. Photosystems I and II both play a role in the cyclic light reactions.

D. O2 is a product of the light-dependent reactions.

14 . If a couple has had three sons and the woman is pregnant with their fourth child, what is the probability that child 4 will also be male?

A. 1⁄2

B. 1⁄4

C. 1⁄8

D. 1⁄16

15 . Which of the following is an incorrect statement about gel electrophoresis?

A. DNA migrates from positive charge to negative charge.

B. Smaller DNA travels faster.

C. The DNA migrates only when the current is running.

D. The longer the current is running, the farther the DNA will travel.

16 . You are told that in a population of guinea pigs, 4 percent are black (recessive) and 96 percent are brown. Which of the following is the frequency of the heterozygous condition?

A. 16 percent

B. 32 percent

C. 40 percent

D. 48 percent

17 . Which of the following is known to be involved in the photoperiodic flowering response of angiosperms?

A. Auxin

B. Cytochrome

C. Phytochrome

D. Gibberellins

18 . Which of the following tends to be highest on the trophic pyramid?

A. Primary consumers

B. Herbivores

C. Primary carnivores

D. Primary producers

19 . A form of species interaction in which one of the species benefits while the other is unaffected is called

A. parasitism.

B. mutualism.

C. commensalism.

D. symbiosis.

20 . The transfer of DNA between two bacterial cells connected by sex pili is known as

A. specialized transduction.

B. conjugation.

C. transformation.

D. generalized transduction.


For questions 21–22, please use the preceding diagram:

21 . If inhibitor 1 is able to bind to the active site and block the attachment of the substrate to the enzyme, this is an example of

A. noncompetitive inhibition.

B. competitive inhibition.

C. a cofactor.

D. a coenzyme.

22 . Which of the following is not a change that would affect the efficiency of the enzyme shown above?

A. Change in temperature

B. Change in pH

C. Change in salinity

D. Increase in the concentration of the enzyme


23 . Which of the following points on the preceding energy chart represents the activation energy of the reaction involving the enzyme?

A. A

B. B

C. C

D. D

For questions 24–27, please use the following answers:

A. Aposomatic coloration

B. Batesian mimicry

C. Deceptive markings

D. Cryptic coloration

24 . Those being hunted adopt a coloring scheme that allows them to blend in to the colors of the environment.

25 . An animal that is harmless copies the appearance of an animal that is dangerous as a defense mechanism to make predators think twice about attacking.

26 . Warning coloration adopted by animals that possess a chemical defense mechanism.

27 . Some animals have patterns that can cause a predator to think twice before attacking.

Questions 28–31 refer to the following choices:

A. Enhancer

B. Repressor

C. Operator

D. Promoter

28 . Short sequence by promoter that assists transcription by interacting with regulatory proteins.

29 . Protein that prevents the binding of RNA polymerase to the promoter site.

30 . Transcription-affecting DNA region that may be located thousands of basepairs away from the promoter.

31 . Basepair sequence that signals the start site for gene transcription.

Questions 32–35 refer to the following choices:

A. Divergent evolution

B. Convergent evolution

C. Parallel evolution

D. Coevolution

32 . Two unrelated species evolve in a way that makes them more similar.

33 . Similar evolutionary changes occurring in two species that can be related or unrelated.

34 . The tandem back-and-forth evolution of closely related species, which is exemplified by predator-prey relationships.

35 . Two related species evolve in a way that makes them less similar.


Questions 36–39 refer to the preceding pedigree.

36 . What kind of inheritable condition does this pedigree appear to show?

A. Autosomal dominant

B. Autosomal recessive

C. Sex-linked dominant

D. Sex-linked recessive

37 . What is the probability that couple C and D will produce a child that has the condition?

A. 0

B. 0.125

C. 0.250

D. 0.333

38 . Which of the following conditions could show the same kind of pedigree results?

A. Cri-du-chat syndrome

B. Turner syndrome

C. Albinism

D. Hemophilia

39 . If child E does in fact have the condition, what is the probability that child F will also have it?

A. 0

B. .250

C. .500

D. .750

Questions 40–42: An experiment involving fruit flies produced the following results:

Vestigial wings are wild type, crumpled wings are mutant.

Gray body is dominant, black body is mutant.


40 . From the data presented above, one can conclude that these genes are

A. sex-linked.

B. epistatic.

C. holandric.

D. linked.

41 . What is the crossover frequency of these genes?

A. 10 percent

B. 20 percent

C. 30 percent

D. 35 percent

42 . How many map units apart would these genes be on a linkage map?

A. 5 map units

B. 10 map units

C. 20 map units

D. 30 map units

Questions 43–45: A laboratory procedure involving plants presents you with the data found in the following 2 charts:


Transpiration Rate1.0 = Control Rate (All Leaves Have the Same Surface Area)


43 . From the transpiration rate data, it appears that transpiration rate rises as

A. temperature ↑, wind speed ↓, humidity ↓

B. temperature ↑, wind speed ↑, humidity ↓

C. temperature ↑, wind speed ↑, humidity ↑

D. temperature ↓, wind speed ↑, humidity ↑

44 . According to the Rf values given in the preceding smaller table, which pigment would migrate the fastest on chromatography paper?

A. Xanthophyll

B. Chlorophyll a

C. Chlorophyll b

D. Beta carotene

45 . From the transpiration rate data presented in the preceding larger table, which of the following plants appears to be most resistant to transpiration?

A. Plant A

B. Plant B

C. Plant C

D. Plants B and C are similarly resistant

Questions 46–48: A population of rodents is studied over the course of 100 generations to examine changes in dental enamel thickness. Species that are adapted to eat food resources that require high levels of processing have thicker enamel than do those that eat softer, more easily processed foods. Answer the following questions using this information and the curves that follow.

46 . How is average enamel thickness changing in this population?

A. There is no real change.

B. The color and size are changing.

C. It is increasing.

D. It is decreasing.

47 . You randomly pick one data point from all three sets of data (all three generations), and the individual’s enamel thickness score is 15. Which of the following can be inferred?

A. The individual comes from generation 1.

B. The individual comes from generation 50.

C. The individual comes from generation 100.

D. The individual could be from any of these generations.

48 . What inference can you make about this species’ diet?

A. Its food resources are getting softer and easier to process.

B. Its food resources are getting harder and more difficult to process.

C. The population is growing.

D. The population is shrinking.


Questions 49–52: A student sets up a lab experiment to study the behavior of slugs. She sets up a large tray filled with soil that measures 1 square meter and has four sets of conditions, one in each quadrant:


She places 20 slugs in the tray, 5 in each quadrant. Use this information to answer the following questions:

49 . What is this lab setup called?.

A. A gel sheet

B. A choice chamber

C. A potometer

D. An incubation chamber

50 . After 5 minutes, there are 5 slugs in each quadrant. Which of the following is not a viable explanation for this finding?

A. The slugs haven’t had time to move yet.

B. The slugs have no preference for temperature or salinity conditions.

C. The slugs can’t move from one area of the tray to another.

D. The slugs do not like to live in high-temperature areas.

51 . After 20 minutes, 20 slugs are in the high-temperature, low-salinity quadrant. What kind of animal behavior has this experiment displayed?

A. Kinesis

B. Taxis

C. Survival

D. Feeding

52 . A classmate has set up a similar experiment in the following manner:


Of the 20 slugs that she puts in her tray, 18 move to the high-salinity, high-temperature section within one hour, while the other 2 move to the low-salinity, low-temperature section. She concludes that slugs prefer conditions of high salinity and temperature. What is wrong with this conclusion?

A. She didn’t specify what the two temperatures or salinities were.

B. The slugs may not have been able to move where they wanted.

C. Crowding may have affected the behavior of the slugs, causing the 2 others to move to the other section.

D. She is measuring two variables at once with no control, and therefore can’t conclude anything about slug tastes.

53 . Viral transduction is the process by which viruses carry bacterial DNA from one bacterial cell to another. In what way does this process play a role in bacterial evolution?

A. By making the bacterial cell more resistant to predators

B. By directly creating new species of bacteria

C. By increasing genetic variation of the bacteria

D. By selecting for viruses better able to infect bacteria

54 . ADH is a hormone secreted by the kidneys that reduces the amount of water excreted in the urine. ADH is released in times of dehydration. This is an example of

A. innate behavior.

B. maintaining homeostasis.

C. failure to respond to the environment.

D. positive feedback.

For questions 55–57 refer to the information and graph that follows.

Five dialysis bags, made from a semipermeable membrane that is impermeable to glucose, were filled with various concentrations of glucose and placed in separate beakers containing 0.5 M glucose solution. The bags were weighed every 10 minutes and the percent change in mass for each bag was graphed:


55 . Which line represents the bag that contained a solution isotonic to the 0.5 M solution?

A. A

B. B

C. C

D. D

56 . Which line represents that bag with the highest initial concentration of glucose?

A. A

B. B

C. C

D. D

57 . Which line or lines represent bags that contain a solution that is hypertonic at 50 minutes?

A. A and B

B. B

C. C

D. D and E

58 . A mutation in a bacterial enzyme changed a previously polar amino acid into a nonpolar amino acid. This amino acid was located at a site distant from the enzyme’s active site. How might this mutation alter the enzyme’s substrate specificity?

A. By changing the enzyme’s pH optimum

B. By changing the enzyme’s location in the cell

C. By changing the shape of the protein

D. An amino acid change away from the active site cannot alter the enzyme’s substrate specificity.

Use the following picture of DNA to answer questions 59 and 60:


59 . Based on the preceding picture, which direction would RNA polymerase move?

A. 3′ → 5′ along the template strand

B. 3′ → 5′ along the complementary strand

C. 5′ → 3′ along the template strand

D. 5′ → 3′ along the complementary strand

60 . If the DNA segment is a transcriptional unit, where would the promoter be located?

A. To the left of the complementary strand

B. To the right of the template strand

C. To the left of the template strand

D. To the right of the complementary strand

61 . A single gene from five related species of leafhopper was compared, and the nucleotide differences between the genes are as shown in the table:

Nucleotide Differences


Which of the following phylogenetic trees best shows the correct evolutionary relationship between the leafhoppers?

A. images

B. images

C. images

D. images

Answer questions 62 and 63 based on the following cladogram:


62 . What is the common ancestor for B and E?

A. 1

B. 2

C. 3

D. 4

63 . Which two species are most closely related?

A. A and E

B. A and B

C. B and C

D. D and E


Calculate the correct answer and enter it on the top line of the grid-in area with each number or symbol in a separate column. Then fill in the correct circle below each number or symbol you entered (only one filled-in circle per column).

1 . Twenty people decide to start a new population, totally isolated from anyone else. Two of the individuals are heterozygous for a recessive allele, which in homozygotes causes cystic fibrosis. Assuming this population is in Hardy-Weinberg equilibrium, what fraction (expressed as a decimal) of people in this new population will have cystic fibrosis?

2 . A certain mutation found in fruit flies (Drosophila melanogaster ) is hypothesized to be autosomal recessive. The experimenter crossed two Drosophila flies that were heterozygous for the trait. The next generation produced 70 wild-type males, 65 wild-type females, 36 males with the mutation, and 40 mutant females. Calculate the chi-squared value for the null hypothesis that the mutation is autosomal recessive.

3 . If the pH of a solution is calculated using the equation pH = −log[H+ ], what is the pH of a solution with a hydrogen ion concentration of 1.33 × 10−8 ?

4 . A cell is in equilibrium with its surroundings. The molarity of the surrounding solution is 0.8 M. Calculate the solute potential of the surrounding solution.

The equation for solute potential is Ψs = −i CRT where

i = ionization constant (assume that it is 1)

C = molar concentration

R = pressure constant (R = 0.00831 liter MPa/mole K)

T = temperature in kelvins (room temperature is 293 K)

5 . Treatment of tomato plants with a growth hormone yielded the following weights of tomatoes: 100 g, 86 g, 123 g, 98 g, 104 g, 71 g. What is the average weight of a tomato after treatment?

6 . After seven days of growth, a plant’s weight was 14.3 grams. The percent biomass of that plant was determined to be 23.1 percent. What amount of energy (in kcal) is stored in the plant, if the amount of stored energy = (g biomass) × 4.35 kcal?

AP Biology Practice Exam 2: Section II


Time—1 hour and 30 minutes

(The first 10 minutes is a reading period. Do not begin writing until the 10-minute period has passed.)
Questions 1 and 2 are long free-response questions that should require about 20 minutes each. Questions 3–8 are short-response questions that should require about 6 minutes each. Outline form is not acceptable. Answers should be in essay form.

1 . The immune system is the body’s defense against foreign invaders and is divided into specific and nonspecific immunity, and humoral and cell-mediated immunity. Answer three of the following four questions:

A. Describe the primary immune response and how an invading antigen is met, dealt with, and eliminated. Describe the cells involved and how they are created.

B. Describe the mechanism by which the immune system deals with viruses, invaders that make it inside our cells .

C. Define nonspecific immunity, and list three examples of nonspecific defense mechanisms in humans.

D. Define the term vaccination , and describe how a vaccination works.

2 . You just started working at a local laboratory and are usually given the grunt-work lab assignments to perform. Design and describe how you would do the following experiments:

A. Describe how you would design an experiment to prove the theory that photosynthesis requires both light and chloroplasts. Describe what equipment you would use, what your control would be, and how your expected outcome would support your hypothesis.

B. You are told that you need to determine how the following factors affect the rate of transpiration in plants: temperature, humidity, light intensity, and air movement. Describe how you would perform an experiment that could accomplish that task, and give your prediction of the expected results. Be sure to describe your experimental setup.

3 . The phenotype for scale color in gila monsters is determined by a specific locus. The dominant allele (black) is represented by G and the recessive allele (brown) is represented by g. The cross between a male gila monster with black scales and a female gila monster with brown scales produced the following F1 generation:

• Black-scaled gila monsters: 52

• Brown-scaled gila monsters: 55

• White-scaled gila monsters: 1

The black-scaled females and brown-scaled males from the F1 generation were then crossed to produce the following F2 generation:

• Black-scaled gila monsters: 53

• Brown-scaled gila monsters: 54

• White-scaled gila monsters: 0

A. Based on the data presented here, determine the P-generation genotypes. Provide Punnett squares that support your answer.

B. The white-scaled female in the F1 generation resulted from a mutational change. Explain what a mutation is and discuss a type of mutation that might have produced the white-scaled female in the F1 generation.

4 . The idea of surface area is an important concept in biology. Explain how surface area plays a critical role in the digestive system.

5 . The following table includes data from scan samples conducted on a fictional mammal called a googabear every 10 minutes over the course of 42 hours. At each scan, it was noted whether the googabear was active or inactive. The percentage of active (feeding, moving, engaging in social behavior) and inactive (resting or sleeping) scans recorded for each time period are shown in the table. Describe the pattern of activity for the googabear and discuss possible reasons for this pattern.


6 . In Earth’s early history, the evolution of photosynthesis in simple cells occurred before the evolution of more complex cells. Briefly describe the significance of photosynthesis being present first.

7 . What evidence supports the theory that chloroplasts and mitochondria are evolved from prokaryotic cells?

8 . You are asked to estimate if a certain species of plant could live in a salt marsh. You collect the following data:

• The overall Ψ of the soil (Ψsoil ): −2.2 MPa

• Solute concentration of plant cell contents: 0.08 M (assume i = 1, and 12 ° C

• Pressure potential of the plant cells: −1.2 MPa

• R = 0.00831 liter MPa/mole K

Do you think the plant could grow in this environment? Why or why not? Show your work.

image Answers and Explanations for Practice Exam 2


1 . D —Habituation is the loss of responsiveness to unimportant stimuli or stimuli that do not provide appropriate feedback. This is a prime example of habituation.

2 . A —Directional selection occurs when members of a population at one end of a spectrum are selected against, while those at the other end are selected for. Taller giraffes are being selected for; shorter giraffes are being selected against.

3 . A —When interbreeding ceases because some sort of barrier separates a single population into two (an area with no food, a mountain, etc.), the two populations evolve independently, and if they change enough, then, even if the barrier is removed, they cannot interbreed. This is allopatric speciation.

4 . DChapter 19 , despite being last, is a very important chapter. The experiments are very well represented on the AP Biology exam, and you should read this chapter carefully and learn how to design and interpret experiments.

5 . C

6 . A

7 . D

8 . B —The endosymbiotic theory proposes that mitochondria and chloroplasts evolved through the symbiotic relationship between prokaryotic organisms.

9 . D —Fibroblast growth factor is said to be involved, but fibroblast cells are not.

10 . D —Posttranscriptional modification actually occurs in the nucleus.

11 . C —Genetic drift is a change in allele frequencies that is due to chance events. When drift dramatically reduces population size, it is called a “bottleneck.”

12 . A —Neutrophils are phagocytic cells of the immune system. They roam the body looking for rubbish to clear.

13 . C —Only photosystem I is involved in the cyclic reactions. Photosystem II is not.

14 . A —Genetics has no memory . . . it will be 1⁄2 forever.

15 . A— DNA migrates from a negative charge to a positive charge. The rest are true.

16 . B —0.04 = q 2 . Therefore, the square root of 0.04 = q = 0.20 and p + q = 1. So p + 0.20 = 1. Therefore, p = 0.80, and 2pq is the frequency of the heterozygote condition: 2(0.20)(0.80) = 0.320 = 32 percent.

17 . C —Phytochrome is an important pigment to the process of flowering. Of its two forms, the active form, Pfr , is responsible for the production of the hormone florigen, which is thought to assist in the blooming of flowers.

18 . C— Secondary carnivores > primary carnivores > primary consumers = herbivores > primary producers.

19 . C —The example to know is the cattle egrets that feast on insects aroused into flight by cattle grazing in the insects’ habitat. The birds benefit because they get food, but the cattle do not appear to benefit at all.

20 . B —Conjugation is the sexual reproduction of bacteria.

21 . B —In competitive inhibition, an inhibitor molecule resembling the substrate binds to the active site and physically blocks the substrate from attaching.

22 . C —The other four are the four main factors that can affect enzyme efficiency.

23 . B —The activation energy of a reaction is the amount of energy needed for the reaction to occur. Notice that the activation energy for the enzymatic reaction is much lower than the nonenzymatic reaction.

24 . D

25 . B

26 . A

27 . C

28 . C

29 . B

30 . A

31 . D

32 . B

33 . C

34 . D

35 . A

36 . B —It is not autosomal dominant because in order for the second generation on the left to have those two individuals with the condition, one parent would need to display the condition as well. It is probably not sex-linked because it seems to appear as often in females as in males. Autosomal recessive seems to be the best fit for this disease.

37 . D —One first needs to determine the probability that person C is heterozygous (Bb). We know that person D is double recessive because she has the condition. We know that the parents for person C must be Bb and Bb because neither of them has the condition, but they produced children with the condition. The probability of person C being heterozygous is 2⁄3, because a monohybrid cross of his parents (Bb × Bb) gives the following Punnett square:


Since you know that he doesn’t have the condition, he cannot be bb. This leaves just three possible outcomes, two of which are Bb. A cross must then be done between the father (person C) Bb and the mother (person D) bb. The chance of their child being bb is 50 percent or 1⁄2. This means that the chance of these two having a child with the condition is 2⁄3 × 1⁄2 or 1⁄3.

38 . C —Albinism is the only autosomal recessive condition on this list.

39 . C —It is 1⁄2, because finding out that one of their children has the condition lets us know that the father (person C) is definitely Bb. This changes the probability of 2⁄3 to 1, meaning that the probability of the two having another child with this condition is simply the result of the Punnett square of Bb × bb, or 1⁄2.

40 . D —When you see a ratio like the one in this problem—7:7:1:1 (approximately)—the genes are probably linked. The reason the crumpled, gray, and vestigial black flies exist at all is because crossover must have occurred.

41 . A —To determine the crossover frequency in a problem like this, simply add up the total number of crossovers (75 + 45 = 120) and divide that sum by the total number of offspring (120 + 555 + 525 = 1200). This results in 120/1200 or 10 percent.

42 . B —One map unit is equal to a 1 percent recombination frequency.

43 . B —The data in the table show you that this answer is the correct choice.

44 . D —The larger the value of Rf for a bunch of pigments dissolved in a particular chromatography solvent, the faster the pigments will migrate. Beta carotene has the highest Rf value.

45 . B —Across the board it seems to have the lowest rate of transpiration. You can make this leap because, as mentioned on top of the larger chart, all the leaves have the same surface area, allowing you to compare their transpiration values.

46 . C —The average enamel thickness started at 10, increased to 12, and then increased to 15. It is therefore increasing overall.

47 . D —The average enamel thickness does not describe the range of possible values; an individual with a thickness of 15 could reasonably come from any of the three generations (if we took into account probability, we could say that the individual most likely came from the 100th generation because this population has the highest frequency of individuals with this thickness; however, the question does not ask for probabilities).

48 . B —Because thicker enamel in this species indicates foods that are more difficult to process, the answer is B. Answer E is incorrect because our model has no predictive power; if the food resources change, the enamel thickness may as well, to either a thicker or thinner average (enamel thickness could also stay the same).

49 . B —Experimental setups where individuals are given a choice as to where to move are called “choice chambers.”

50 . D —All the answers except D are possible, and are important things to consider when setting up an experiment. For example, it is important to allow your study animals enough time to move and/or get used to their new surroundings and conditions before drawing conclusions about their behavior. D is not a good answer because half of the slugs started in a high-temperature area and haven’t moved.

51 . A —Kinesis is the movement of animals in response to current conditions; animals tend to move until they find a favorable environment, at which point their movement slows.

52 . D —It is important to try to measure only one variable at once. The 18 slugs may have moved to the higher-temperature, higher-salinity conditions because they need high temperatures to survive, even if they dislike high salinity, and vice versa. The original experiment circumvents this problem by giving a choice for all the possible combinations of variables. Answer E is an interesting issue, but two individuals is probably too small a number to warrant throwing away the study results.

53 . C —New genes are introduced into the bacterium through viral transduction.

54 . B —When the body has too little water, ADH works to increase the amount of water available. This drive to maintain a stable condition is an example of homeostasis.

55 . C —Line C showed no net change in weight, indicating the concentration of the solution inside the bag was the same (isotonic) as the solution in the beaker.

56 . A —The most water would diffuse into the most hypertonic solution; line A shows the biggest increase in weight.

57 . B —Line B still shows an increase in weight at 50 minutes, whereas line A has leveled out and is isotonic at 50 minutes.

58 . C —Even though an amino acid doesn’t have direct contact with the substrate, it still plays a role in the overall shape of the enzyme.

59 . A —As RNA polymerase adds new nucleotides to the 3′ end of the new strand, it is moving toward the 5′ end of the (antiparallel) template strand.

60 . B —The promoter would be located upstream from where transcription would begin.

61 . C —There are few nucleotide differences between species 1 and 2, indicating they would reside close to one another on the cladogram. The same holds true for species 3 and 5. There are large numbers of differences between species 4 and all others, indicating it would be positioned on its own branch.

62 . A —Both B and E branches originate from point 1.

63 . C —Species B and C reside the closest to one another.


1 . .0025 —For this specific gene in this specific population, there are a total of 40 alleles, two of which are the recessive cf allele (2/40 = 0.05 = q ). Since you need to be homozygous recessive to have cystic fibrosis, (q ) × (q ) = q 2 = (0.05)2 = .0025. In other words, 25 out of 10,000 people (0.25 percent) will have cystic fibrosis.


2 . 13.33 —If both parents were heterozygous and if this trait is indeed recessive, you would expect the next generation to show 75 percent normal-looking flies and 25 percent of the flies with the recessive trait. Based on a total of 211 flies, that would mean you would expect 158 normal flies and 53 recessive flies. Your observed numbers were, instead, 135 normal flies and 76 recessive flies.


Since your chi-squared value (13.33) is higher than the critical value of 6.64 (based on 1 degree of freedom), you have to reject your hypothesis. Something other than an autosomal recessive trait is going on.


3 . 7.88 —pH = −log(1.33 × 10−8 ) = 7.88


4 . –1.95 —Ψs = −(1)(0.8 M)(0.00831 L bars/mole K)(294 K) = –1.95 MPa


5 . 97 —Add the weight of all tomatoes and divide by the number of tomatoes. The average is 97 grams.


6 . 14.4 —The computation is shown below:

(14.3 g)(0.231) = 3.30 g biomass

(3.30 g biomass)(4.35 kcal) = 14.4 kcal


image Free-Response Grading Outline

1 . Immune system question (here, the student can obtain 4 points from two of the answers; if 4 points are awarded for an answer, a maximum of 3 points can be obtained for each of the remaining answers)

A. (maximum 4 points)

• Definition of an antigen as a molecule foreign to the body. (1⁄2 point)

• Mentioning that the primary immune response is an example of humoral immunity. (1⁄2 point)

• Description of a B cell and how each B cell has a specific antigen recognition site on its surface that will match up with only one antigen. (1 point)

• When B cells meet and attach to the appropriate antigen, they become activated and undergo mitosis and differentiation into two types of cell. (1 point)

• The two types of cell are the memory cells and plasma cells. (1⁄2 point)

• Definition of plasma cells as the cells that produce the specific antibodies. (1⁄2 point)

• Definition of memory cells as the cells that head up the secondary immune response. (1⁄2 point)

• Description of how an antibody recognizes a particular antigen, including the fact that antibodies have two functional regions: Fab , which binds to the antigen; and Fc , which binds to the effector cells, and later comes in and cleans up the trash left behind. (1 point)

• Mentioning that complement is the one that binds to the antigen–antibody complex and aids in the quicker removal of the complex from the body. (1⁄2 point)

B. (maximum 3 points)

• Mentioning that this portion of the immune system is known as cell -mediated immunity . (1⁄2 point)

• Mentioning that the major player involved here is the cytotoxic T cell. (1 point)

• Mentioning that the cells infected by a virus are forced to produce viral antigens, some of which show up on the surface of the cell, and that it is these antigens that cytotoxic T cells recognize and attack. (1 point)

• Mentioning that all cells of the human body (except red blood cells) have class I histocompatibility antigens (MHC I) on their surfaces. (1 point)

• Further discussion of MHC—mentioning that MHC I antigens are slightly different for each person and the immune system accepts any cell that has the identical MHC I as friendly, and any cell that has a different form of MHC on its surface as an enemy. (1 point)

C. (maximum 4 points)

• Definition of nonspecific immunity as the nonspecific prevention of the entrance of invaders into the human body. (1 point)

• Examples (each example is worth 1 point)

a. Lysozyme in the saliva can kill germs before they have the chance to take hold.

b. The skin covering the body is a major form of nonspecific protection against invasion.

c. The mucous membrane lining the trachea and lungs prevent bacteria from entering cells and actually assist in the expulsion of bacteria by ushering them up and out with a cough.

d. The low pH of the stomach (acidity) is a nonspecific defense mechanism because it is able to kill a lot of bacteria that enter the body that cannot handle such an acidic environment.

D. (maximum 4 points)

• Definition of a vaccination as something given to an individual in an effort to prime the immune system to be prepared to fight a specific sickness if confronted again in the future. (1 point)

• Recognition that a vaccination is the injection of an antigen into the system (human body). (1⁄2 point)

• Description of how the reception of an antigen by a B cell causes B-cell differentiation into memory and plasma cells. (1⁄2 point)

• Mentioning that at the time of the vaccination, the plasma cells will produce antibodies to wipe out the small dose of antigen presented during the vaccination, and that the memory cells will remember the antigen and be ready to react later if necessary. (1 point)

• Definition of a secondary immune response. Memory cells are stored instructions on how to handle a particular invader. When the invader returns to the body, the memory cells recognize it and produce antibodies in rapid fashion. (1 point)

• Mentioning that the secondary immune response is faster and more efficient than the primary immune response. (1⁄2 point)

• Mentioning that the principle of a successful vaccination rests on the belief that the secondary immune response will succeed and wipe out the sickness if the individual is exposed in the future. (1⁄2 point)

2 . Plant laboratory question

A. (maximum 5 points)

• Mentioning that the products of the light reactions of photosynthesis are ATP, NADPH, and oxygen. (1⁄2 point)

• Mentioning that in this experiment, the NADP+ would be replaced by a compound known as DPIP. (1⁄2 point)

• Mentioning that normally this compound DPIP has a nice blue color, but when reduced, it changes to a colorless solution. (1⁄2 point)

• Mentioning that a machine called a spectrophotometer will be used to measure the amount of light that can pass through various samples. (1⁄2 point)

• Description of the experiment.

a. Set aside three beakers—one with boiled chloroplasts, two with unboiled chloroplasts. (1 point)

b. Take initial reading on spectrophotometer to determine how much light passes through the unboiled chloroplasts before the experiment begins. (1⁄2 point)

c. Take one sample (unboiled chloroplasts) and measure how much photosynthesis occurs while it sits in a dark environment. After a certain amount of time, use the spectrophotometer to measure how much light can pass through the solution. (1 point)

d. Take a second sample (unboiled chloroplasts) and measure how much photosynthesis occurs when it is exposed to light. After a certain amount of time, use the spectrophotometer to measure how much light can pass through the solution. (1 point)

• Mentioning that they would now compare the two samples to see the effect of light on photosynthesis. (1 point)

• Take a third sample (boiled chloroplasts) and expose it to light, and after a certain period of time, measure how much light can pass through the solution. (1 point)

• Mentioning that they would now compare the third sample and the second sample to see the effect of the presence or absence of chloroplasts on photosynthesis. (1 point)

B. (maximum 5 points)

• Definition of transpiration as the evaporative water loss from plants. (1 point)

• Mentioning that they will use a potometer to measure the amount of water loss from plants. (1 point)

• Mentioning that the surface area of a leaf is important to the measurement of transpiration rate in an experiment of this nature. (1⁄2 point)

• Description of experiment.

a. Begin by measuring the amount of water that evaporates from the surface of a plant over a certain amount of time under normal conditions. Use this as the control. (1 point)

b. Change the temperature, humidity, light intensity, and air movement that the plant is exposed to by 5-degree increments, and measure the amount of transpiration that occurs at the various temperatures. (1 point for each variable mentioned up to a maximum of 2 points)

c. Mention that these values will be compared to the control to determine the effect of temperature, humidity, light intensity, and air movement. (1⁄2 point for each variable mentioned up to a maximum of 1 point)

• Mentioning that transpiration increases with an increase in temperature, decrease in humidity, increase in light intensity, and increased air movement. (1 point for each, up to a maximum of 2 points)

3 . The phenotype for scale color in gila monsters is determined by a specific locus. (maximum 4 points for entire question)

A. P-generation genotypes (maximum 2 points)

• Creating the correct Punnett square, as shown here (1 point):


• Mentioning that genotype of P-generation = Gg (heterozygous) crossed with gg (homozygous recessive). (1 point)

B. Explanation for mutation (maximum 3 points)

• Mentioning that mutation is a random event that causes change in allele frequencies. (1 point)

• Possible explanation: scale-color gene is tied to another gene that controls pigment distribution (gene at one locus alters phenotypic expression of a gene at another locus). (1 point)

• Using the term epistasis correctly. (1 point)

• Possible explanation: point mutation occurred as DNA responsible for the production of protein that determines scale color was undergoing replication. (1 point)

4 . The idea of surface area is an important concept in biology. Explain how surface area plays a critical role in the digestive system. (maximum 4 points for entire question)

• Mentioning that majority of absorption occurs in small intestine. (1 point)

• Mentioning that numerous folds and ridges increase surface area. (1 point)

• Mentioning that brush border and microvilli increase surface area. (1 point)

• Mentioning that large surface area leads to greater absorption of nutrients. (1 point)

• Mentioning that chewing (mastication) breaks up food into smaller pieces. (1 point)

• Mentioning that higher surface area gives greater access to salivary amylase. (1 point)

5 . The following table includes data from scan samples conducted on a fictional mammal called a googabear every 10 minutes over the course of 42 hours. (maximum 4 points for entire question)

• Mentioning that googabears are more active from 6 a.m. until 6 p.m. (during daylight hours). (1 point)

• Mentioning that they decrease activity from 6 p.m. to 6 a.m. (when it is dark). (1 point)

• Mentioning that googabears’ food source is available during daylight. (1 point)

• Mentioning that googabear predators are nocturnal (out at night), so it is safest for googabears to remain hidden at night. (1 point)

• Mentioning that googabears rely on collective body heat at night (huddling); activity of huddled group is low. (1 point)

6 . In Earth’s early history, the evolution of photosynthesis in simple cells occurred before the evolution of more complex cells. (maximum 4 points for entire question)

• Mentioning that photosynthesis releases oxygen as a by-product. (1 point)

• Mentioning that photosynthesis led to an increase in atmospheric oxygen. (1 point)

• Mentioning that the presence of more oxygen in the atmosphere allowed the evolution of cellular respiration. (1 point)

• Mentioning that oxygen allowed cells to generate more energy and grow larger and more complex. (1 point)

• Mentioning that the first photosynthetic cells were prokaryotic. (1 point)

• Mentioning that eukaryotic cells could not evolve until there was a higher level of atmospheric oxygen. (1 point)

7 . What evidence supports the theory that chloroplasts and mitochondria are evolved from prokaryotic cells? (maximum 4 points for entire question)

• Mentioning that chloroplasts and mitochondria have their own DNA. (1 point)

• Mentioning that chloroplast and mitochondrial DNA consist of a single, circular molecule (like bacterial DNA). (1 point)

• Mentioning that chloroplast and mitochondrial DNA are not associated with histones (like bacterial DNA). (1 point)

• Mentioning that chloroplasts and mitochondria replicate by a process similar to prokaryotes. (1 point)

• Mentioning that the inner membranes of both organelles have enzymes homologous to those found in prokaryotes. (1 point)

8 . You are asked to estimate if a certain species of plant could live in a salt marsh. (maximum 4 points for entire question)

• Calculate Ψplant cell :

Ψ = pressure potential + solute potential solute potential of plant cell = Ψs = –iCRT = –(1)(0.08 M)(.00831)(273 + 12) = –0.189 MPa (1 point)

Ψ = –1.2 MPa (pressure potential) + –0.189 MPa (solute potential) = –1.39 MPa (1 point)

• Mentioning that the plant cell’s water potential (–1.39 MPa) is higher than that of the soil (–2.2 MPa). (1 point)

• Mentioning that water would flow out of the plant cell (hypotonic) into soil (hypertonic). (1 point)

• Mentioning that the plant cell would not survive. (1 point)

Scoring and Interpretation


Free-Response Questions:

1. ____ / 10

2. ____ / 10

3. ____ / 4

4. ____ / 4

5. ____ / 4

6. ____ / 4

7. ____ / 4

8. ____ / 4

Add up the total points accumulated in the eight questions and multiply the sum by 1.57 to obtain the free-response raw score: images


Now combine the raw scores from the multiple-choice and free-response sections to obtain your new raw score for the entire practice exam. Use the ranges listed below to determine your grade for this exam. Don’t worry about how we arrived at the following ranges, and remember that they are rough estimates on questions that are not actual AP exam questions . . . do not read too much into them.



Futuyma, Douglas J. Evolutionary Biology . 3d ed. Sunderland, MA: Sinauer Associates, Inc., 1998.

Kotz, John C., Paul M. Treichel, and John Townsend. Chemistry and Chemical Reactivity . 8th ed. Stamford, CT: Brooks/Cole, 2011.

Reece, Jane B., Lisa A. Urry, and Michael L. Cain. Campbell Biology . 9th ed. San Francisco: Benjamin Cummings, 2011.

Starr, Cecie, Christine Evers, and Lisa Starr. Biology: Concepts and Applications . 8th ed. San Francisco: Brooks/Cole, 2011.

Strauss, Eric, and Marylin Lisowski. Biology: The Web of Life . Teacher’s ed. Menlo Park, CA: Addison-Wesley Longman, 1998.

Wilbraham, Antony C. Chemistry . 5th ed. New York: Pearson, 2000.


Here is a list of websites that contain information and links that you might find useful to your preparation for the AP Biology exam: